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Chapter 12: Problem 1769

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (B) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$ (C) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (D) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$

Short Answer

Expert verified
The equivalent resistivity of the combination of two wires of equal diameters and resistivities \(\rho_{1}\) and \(\rho_{2}\) joined in series is: \(\rho_{eq} = \frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{\ell_{1}+\ell_{2}}\).

Step by step solution

01

Expressing the resistance of each wire

The resistance of a wire is given by: \[R = \frac{\rho L}{A}\] Where \(R\) is resistance, \(\rho\) is resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Here, we have two wires with equal diameters and resistivities \(\rho_{1}\) and \(\rho_{2}\), and lengths \(\ell_{1}\) and \(\ell_{2}\), with cross-sectional areas \(A_{1}\) and \(A_{2}\). Since the diameters are equal, their cross-sectional areas are also equal. Thus, we have \(A_{1} = A_{2}\).
02

Finding the equivalent resistance

If these two resistors are connected in series, their equivalent resistance is given by the sum of the individual resistances: \[R_{eq} = R_1 + R_2\]
03

Applying the resistance equation

Now, we apply the resistance formula for each wire, having \(R_1 = \frac{\rho_{1} \ell_{1}}{A_1}\) and \(R_2 = \frac{\rho_{2} \ell_{2}}{A_2}\): \[R_{eq} = \frac{\rho_{1} \ell_{1}}{A_1} + \frac{\rho_{2} \ell_{2}}{A_2}\] Since \(A_1 = A_2\), we can express the equivalent resistance as: \[R_{eq} = \frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{A}\]
04

Determining the equivalent resistivity

We know that the total length of the combination is \(\ell_{1}+\ell_{2}\). Similarly, we can express the equivalent resistance as the product of the equivalent resistivity, \(\rho_{eq}\), and total length of the combination, divided by the cross-sectional area, as: \[R_{eq} = \frac{\rho_{eq} (\ell_{1}+\ell_{2})}{A}\]
05

Comparing the equivalent resistance expressions

Comparing the expressions from Step 3 and Step 4: \[\frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{A} = \frac{\rho_{eq} (\ell_{1}+\ell_{2})}{A}\] By simplifying and isolating the equivalent resistivity \(\rho_{eq}\): \[\rho_{eq} = \frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{\ell_{1}+\ell_{2}}\] Our answer corresponds to option (A) in the given choices.

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Most popular questions from this chapter

In each of the following questions, match column \(\mathrm{I}\) and column II and select the correct match out of the four given choices Column I \(\quad\) Column II (a) The unit of electrical resistivity is (p) \(\mathrm{m}^{2} \mathrm{~S}^{-1} \mathrm{~V}^{-1}\) (b) The unit of current density is (q) \(\Omega^{-1} \mathrm{~m}^{-1}\) (c) The unit of electrical conductivity is (r) \(\mathrm{Am}^{-2}\) (d) The unit of electric mobility is (s) \(\Omega \mathrm{m}\) (A) \(a-p, b-q, c-r, d-s\) (B) \(a-s, b-r, c-q, d-p\) (C) \(a-r, b-q, c-p, d-s\) (D) \(a-q, b-r, c-s, d-p\)

An electrical circuit is shown in fig the values of resistances and the directions of the currents are shown A voltmeter of resistance \(400 \Omega\) is connected across the \(400 \Omega\) resistor. The battery has negligible internal resistance.The residing of the voltmeter is. \((\mathrm{A})(10 / 3) \mathrm{V}\) (B) \(5 \mathrm{~V}\) (C) \((20 / 3) \mathrm{V}\) (D) \(4 \mathrm{~V}\)

Which of the following set up can be used to verify the ohm's law?

A circuit with an infinite no of resistance is shown in fig. the resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\), when $\mathrm{R}_{1}=1 \Omega\( and \)\mathrm{R}_{2}=2 \Omega$ will be (A) \(4 \Omega\) (B) \(1 \Omega\) (C) \(2 \Omega\) (D) \(3 \Omega\)

Match the physical quantities given in column I with their dimensional formulae given in column II - I stands for the dimension of current. \(\begin{array}{ll}\text { Column I } & \text { Column II }\end{array}\) (a) Electromotive force (emf) (p) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (b) Resistance (q) \(\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (c) Resistivity (r) \(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) Conductivity (s) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\) (A) \(a-s, b-p, c-q, d-r\) (B) \(a-p, b-s, c-r, d-p\) (C) \(a-p, b-s, c-r, d-q\) (D) \(a-r, b-p, c-q, d-s\)
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